What is the solution for y^'= 1/(ty) ?

The solution for y^'= 1/(ty) is y=sqrt(2ln(t)+c_{1)},y=-sqrt(2ln(t)+c_{1)}

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y^'= 1/(ty)

y^{^{'}} = \frac{1}{t y}

y = 2 ln (t) + c_{1}, y = - 2 ln (t) + c_{1}

Solution steps

y^{'} (t) = \frac{1}{t y}

Solve separable ODE:

y = 2 ln (t) + c_{1}, y = - 2 ln (t) + c_{1}

y = 2 ln (t) + c_{1}, y = - 2 ln (t) + c_{1}

g (x) = \frac{1}{5 x ^{2} + 6 x}

\frac{\partial}{\partial x} (\frac{1}{y} cos (\frac{x}{y}))

θ \to 0 lim (\frac{1 - cos ( θ )}{sin ( θ )})

\int sec (x) tan (x) - 7 e^{x} d x

x \to 6 lim (3)

What is the solution for y^'= 1/(ty) ?
The solution for y^'= 1/(ty) is y=sqrt(2ln(t)+c_{1)},y=-sqrt(2ln(t)+c_{1)}